Question

a particle with charge q is to be brought from far away to a point near an electric dipole. no work is done if the final position of the particle is on the line through the charges of the dipole. a line that makes an angle of 45° with the dipole moment. a line that makes an angle of 30° with the dipole moment. a line that is perpendicular to the dipole moment. none of the choices are correct.

Explanation:

Work done in moving a charge in an electric - field is given by \(W = q\Delta V\). The electric potential \(V\) due to an electric dipole at a point is \(V=\frac{k\vec{p}\cdot\vec{r}}{r^{3}}\) (where \(\vec{p}\) is the dipole - moment vector and \(\vec{r}\) is the position vector of the point from the dipole). No work is done when the initial and final potentials are the same. The potential due to a dipole is constant on an equipotential surface. The equipotential surfaces of a dipole are the surfaces where the angle between the dipole - moment vector and the position vector of the point from the dipole is \(90^{\circ}\). When the particle is moved to a point on a line perpendicular to the dipole moment, the potential at the initial (far - away, \(V_i = 0\) as \(r
ightarrow\infty\)) and final positions is the same (since on the perpendicular bisector of the dipole, the potential due to the dipole is zero).

Answer:

A. a line that is perpendicular to the dipole moment.